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For example, let us verify the first of the formulas ( We have to
show that, for every function fe ^C(G)? tne continuous function (x,y) ^f(xy)
is (gs®ju)-integrable. Now the function x\-+f(xy) is es-integrable, and

[f(xy)des(x) =f(sy). Since the function y*-*f(sy) is continuous and has
compact support, it is ju-integrable. Our assertion therefore follows from
"(13.21.10), and then the theorem of Lebesgue-Fubini (13.21.7) gives
\[f(xy) dss(x) dn(y) = \f(sy) dp(y). The formula to be proved follows now
from (14.1.1) and (14.1.2).

(14.6.2)   Every finite family (//1?..., jj,n) of bounded measures on G is con-
volvable. The measure /^ * n2 * " " * f*n *s bounded, and

(            llA«i*A«2*'"*A.II ^ llA*ill ' llMill •"llA.II-

To prove the first assertion, observe that the measure /^ ® • • • ® /zrt
on G" is bounded (13.21.18). For every function /e Jfc(G), the function
(xl9..., xn) \-^f(xix2 "• xn) is continuous and bounded on Gn, and therefore
(/*! ® • • • ® /ij-integrable (13.20.4). To prove (, it is enough to
remark that if/e ^TC(G) and ||/|| ^ 1, then

! ll/*iII ' 1102II •" II0J

by virtue of (13.21.18).

(14.6.3)   ^4 /e// /faar measure A on G is conceivable on the right with any
bounded measure n on G, and \JL * A =

We may restrict ourselves to the case where /f g; 0 (cf. (14.7.1 .2)). For each
function /€ «^f+(G) we have

J d/i(*) J

and therefore, by virtue of (13.21.9), the function (x,y)\-*f(xy) is
integrable and its integral is equal to A(/) ||/i||.

The same calculation shows that X is not convotvable with itself if Q is not
compact, for then the function 1 is not A-integrable (14.2.3).

(14.6.4) Let (fa, . . . , /*„) be a finite sequence of measures on G, all of which
except possibly for one have compact support. Then the sequence (jux, . . . , JUM)
is convolvable.
(Problem 6(c)). A subset A of Nc is relatively compact in Nc if and only if it satisfies